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Question Number 153162 by mathdanisur last updated on 05/Sep/21

Determine all the derivable functions f:R→R which satisfy ((f(x^3 ) - f(0))/(f(x) - f(0))) = x^2 ; ∀x≠0

$$\mathrm{Determine}\:\mathrm{all}\:\mathrm{the}\:\mathrm{derivable}\:\mathrm{functions} \\ $$$$\mathrm{f}:\mathbb{R}\rightarrow\mathbb{R}\:\:\mathrm{which}\:\mathrm{satisfy} \\ $$$$\frac{\mathrm{f}\left(\mathrm{x}^{\mathrm{3}} \right)\:-\:\mathrm{f}\left(\mathrm{0}\right)}{\mathrm{f}\left(\mathrm{x}\right)\:-\:\mathrm{f}\left(\mathrm{0}\right)}\:=\:\mathrm{x}^{\mathrm{2}} \:\:;\:\:\forall\mathrm{x}\neq\mathrm{0} \\ $$

Answered by aleks041103 last updated on 05/Sep/21

let g(x)=f(x)−f(0) g(x^3 )=x^2 g(x)⇒((g(x^3 ))/x^3 )=((g(x))/x) let h(x)=g(x)/x=((f(x)−f(0))/x) h(x^3 )=h(x) by induction h(x)=h(x^3^n )=h(x^3^(−n) ), for ∀n∈N let 1>ε>0. case 1: x∈(0,1) we can always find some n such that 1>a=x^3^(−n) >1−ε Then for ∀x∈(0,1) and any 1>ε>0, exists a number a∈[1−ε,1) such that h(x)=h(a). Since we can take ε to be arbitrairly small, then h(x)=lim_(ε→0) h(1>a>1−ε) Since f(x) is differentiable, then h(x) is too(for x≠0) and so h is continuous ⇒h(x∈(0,1))=lim_(ε→0) h(1>a>1−ε)=h(1). case 2: x∈(1,∞) by analogy, we can always find n such that x^3^(−n) ∈(1,1+ε). Then again h(x>1)=lim_(ε→0) h(1<a<1+ε)=h(1) Conclusion: h(x>0)=h(1)=a=const. Analogously one can prove h(x<0)=h(−1)=b=const. Then f(x)= { ((ax+f(0), x≥0)),((bx+f(0), x<0)) :} But for f(x) to be diff. at x=0 we need a=b. ⇒f(x)=ax+f(0) is the only solution!

$${let}\:{g}\left({x}\right)={f}\left({x}\right)−{f}\left(\mathrm{0}\right) \\ $$$${g}\left({x}^{\mathrm{3}} \right)={x}^{\mathrm{2}} {g}\left({x}\right)\Rightarrow\frac{{g}\left({x}^{\mathrm{3}} \right)}{{x}^{\mathrm{3}} }=\frac{{g}\left({x}\right)}{{x}} \\ $$$${let}\:{h}\left({x}\right)={g}\left({x}\right)/{x}=\frac{{f}\left({x}\right)−{f}\left(\mathrm{0}\right)}{{x}} \\ $$$${h}\left({x}^{\mathrm{3}} \right)={h}\left({x}\right) \\ $$$${by}\:{induction} \\ $$$${h}\left({x}\right)={h}\left({x}^{\mathrm{3}^{{n}} } \right)={h}\left({x}^{\mathrm{3}^{−{n}} } \right),\:{for}\:\forall{n}\in\mathbb{N} \\ $$$${let}\:\mathrm{1}>\varepsilon>\mathrm{0}. \\ $$$$\boldsymbol{{c}}{ase}\:\mathrm{1}:\:{x}\in\left(\mathrm{0},\mathrm{1}\right) \\ $$$${we}\:{can}\:{always}\:{find}\:{some}\:{n}\:{such}\:{that} \\ $$$$\mathrm{1}>{a}={x}^{\mathrm{3}^{−{n}} } >\mathrm{1}−\varepsilon \\ $$$${Then}\:{for}\:\forall{x}\in\left(\mathrm{0},\mathrm{1}\right)\:{and}\:{any}\:\mathrm{1}>\varepsilon>\mathrm{0},\: \\ $$$${exists}\:{a}\:{number}\:{a}\in\left[\mathrm{1}−\varepsilon,\mathrm{1}\right)\:{such}\:{that} \\ $$$${h}\left({x}\right)={h}\left({a}\right). \\ $$$${Since}\:{we}\:{can}\:{take}\:\varepsilon\:{to}\:{be}\:{arbitrairly} \\ $$$${small},\:{then} \\ $$$${h}\left({x}\right)=\underset{\varepsilon\rightarrow\mathrm{0}} {\mathrm{lim}}\:{h}\left(\mathrm{1}>{a}>\mathrm{1}−\varepsilon\right) \\ $$$${Since}\:{f}\left({x}\right)\:{is}\:{differentiable},\:{then}\:{h}\left({x}\right) \\ $$$${is}\:{too}\left({for}\:{x}\neq\mathrm{0}\right)\:{and}\:{so}\:{h}\:{is}\:{continuous} \\ $$$$\Rightarrow{h}\left({x}\in\left(\mathrm{0},\mathrm{1}\right)\right)=\underset{\varepsilon\rightarrow\mathrm{0}} {\mathrm{lim}}\:{h}\left(\mathrm{1}>{a}>\mathrm{1}−\varepsilon\right)={h}\left(\mathrm{1}\right). \\ $$$$\boldsymbol{{case}}\:\mathrm{2}:\:\boldsymbol{{x}}\in\left(\mathrm{1},\infty\right) \\ $$$${by}\:{analogy},\:{we}\:{can}\:{always}\:{find}\:{n}\:{such}\:{that} \\ $$$${x}^{\mathrm{3}^{−{n}} } \in\left(\mathrm{1},\mathrm{1}+\varepsilon\right). \\ $$$${Then}\:{again} \\ $$$${h}\left({x}>\mathrm{1}\right)=\underset{\varepsilon\rightarrow\mathrm{0}} {\mathrm{lim}}\:{h}\left(\mathrm{1}<{a}<\mathrm{1}+\varepsilon\right)={h}\left(\mathrm{1}\right) \\ $$$$\boldsymbol{{Conclusion}}: \\ $$$${h}\left({x}>\mathrm{0}\right)={h}\left(\mathrm{1}\right)={a}={const}. \\ $$$$ \\ $$$${Analogously}\:{one}\:{can}\:{prove} \\ $$$${h}\left({x}<\mathrm{0}\right)={h}\left(−\mathrm{1}\right)={b}={const}. \\ $$$$ \\ $$$${Then} \\ $$$${f}\left({x}\right)=\begin{cases}{{ax}+{f}\left(\mathrm{0}\right),\:{x}\geqslant\mathrm{0}}\\{{bx}+{f}\left(\mathrm{0}\right),\:{x}<\mathrm{0}}\end{cases} \\ $$$${But}\:{for}\:{f}\left({x}\right)\:{to}\:{be}\:{diff}.\:{at}\:{x}=\mathrm{0}\:{we}\:{need} \\ $$$${a}={b}. \\ $$$$\Rightarrow{f}\left({x}\right)={ax}+{f}\left(\mathrm{0}\right)\:{is}\:{the}\:{only}\:{solution}! \\ $$

Commented by mathdanisur last updated on 06/Sep/21

Thank you ser nice

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{ser}\:\mathrm{nice} \\ $$

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